2011 AMC 10B Problem 18

Let $\theta = \angle AMD = \angle CMD$.

Because $AB \parallel DC$ and $MD$ is a transversal, $\angle MDC = \angle AMD = \theta$ (alternate interior angles). So in triangle $MDC$ the angles at $M$ and $D$ are both $\theta$, which makes it isosceles with $CM = CD = 6$.

Now look at right triangle $MBC$: $CB = 3$ and hypotenuse $CM = 6$, so $\sin \angle CMB = \frac{3}{6} = \frac{1}{2}$ and $\angle CMB = 30^\circ$.

The three angles at $M$ along line $AB$ add to $180^\circ$:
$$\theta + \theta + 30^\circ = 180^\circ \quad\Longrightarrow\quad \theta = 75^\circ.$$

The answer is $\boxed{\textbf{(E)}\ 75}$.

An equal-angle condition in a figure with parallel lines almost always produces an isosceles triangle; here it converts the angle condition into the length $CM = 6$, and the given $BC = 3$ then hands you a $30$-$60$-$90$ triangle. Look for the hidden isosceles triangle before reaching for coordinates.

Coordinates: $A = (0,0)$, $B = (6,0)$, $C = (6,3)$, $D = (0,3)$, $M = (m, 0)$. The bisector condition $CM = CD$ gives $(6 - m)^2 + 9 = 36$, so $m = 6 - 3\sqrt{3}$; then $\tan \angle AMD = \frac{3}{m} = \frac{3}{6 - 3\sqrt{3}} = 2 + \sqrt{3} = \tan 75^\circ$.

• Guessing $45$ from the picture or assuming $M$ is the midpoint of $AB$, which does not make the two angles at $M$ equal.
• Finding $\angle CMB = 30^\circ$ and answering $30$ or $60$ instead of finishing with $(180 - 30)/2$.